Richard Heathfield's setbits()

Discussion in 'C++' started by Alex Vinokur, May 2, 2010.

  1. Alex Vinokur

    Alex Vinokur Guest

    Hi,

    Here is Richard Heathfield's function from http://users.powernet.co.uk/eton/kandr2/krx206.html

    #include <stdio.h>
    unsigned setbits(unsigned x, int p, int n, unsigned y)
    {
    return (x & ((~0 << (p + 1)) | (~(~0 << (p + 1 - n))))) | ((y & ~(~0
    << n)) << (p + 1 - n));
    }

    setbits(x,p,n,y) returns x with the n bits that begin at position p
    set to the rightmost n bits of y, leaving the other bits unchanged.

    It seems that setbits (x, 31, n, y) may produce undefined behavior.

    According to the C++ standard:
    Behavior of shift operators << and >> is undefined if the right
    operand is negative, or
    greater than or equal to the length in bits of the promoted left
    operand.

    So, result ((~0 << (p + 1)) may be undefined.

    Regards,

    Alex Vinokur
    Alex Vinokur, May 2, 2010
    #1
    1. Advertising

  2. Alex Vinokur <> writes:

    > Here is Richard Heathfield's function from http://users.powernet.co.uk/eton/kandr2/krx206.html
    >
    > #include <stdio.h>
    > unsigned setbits(unsigned x, int p, int n, unsigned y)
    > {
    > return (x & ((~0 << (p + 1)) | (~(~0 << (p + 1 - n))))) | ((y & ~(~0
    > << n)) << (p + 1 - n));
    > }
    >
    > setbits(x,p,n,y) returns x with the n bits that begin at position p
    > set to the rightmost n bits of y, leaving the other bits unchanged.
    >
    > It seems that setbits (x, 31, n, y) may produce undefined behavior.


    Yes, it does.

    > According to the C++ standard:


    K&R is about C so you should probably quote the C standard. The intent
    is that C and C++ remain synchronised about this sort of thing so won't
    matter much, but a solution to a K&R exercise must be assumed to be C.
    As it happens, modern C (C99) has diverged from C++ in the case of left
    shifting negative numbers but the above is, presumably, not C99.

    > Behavior of shift operators << and >> is undefined if the right
    > operand is negative, or
    > greater than or equal to the length in bits of the promoted left
    > operand.
    >
    > So, result ((~0 << (p + 1)) may be undefined.


    It can be undefined for other reasons, though none that matter in the
    context of K&R2. ~0 is often signed and negative in which case it is
    undefined in C99 but not in C90 or in C++ (at least up to and including
    2003). It's safer to shift unsigned types so I'd suggest ~0u.

    I think it's hard to do this without knowing the width of the type. I'd
    probably write:

    unsigned width = CHAR_BIT * sizeof x;
    unsigned mask = ~0u >> width - n << p - n + 1;
    return x & ~mask | y & mask;

    This goes wrong if there are padding bits, but at least we can check for
    that (UINT_MAX will be == ~0u if there are none).

    There is a bit-twiddling version that is one operation shorter:

    return x ^ (x ^ a) & mask;

    but you'd need to justify the "eh?" this might prompt in the reader.

    --
    Ben.
    Ben Bacarisse, May 2, 2010
    #2
    1. Advertising

  3. Alex Vinokur

    Jonathan Lee Guest

    On May 2, 7:46 am, Alex Vinokur <> wrote:
    > Here is Richard Heathfield's function
    > ...
    > setbits(x,p,n,y) returns x with the n bits that begin at position p
    > set to the rightmost n bits of y, leaving the other bits unchanged.
    >
    > It seems that setbits (x, 31, n, y) may produce undefined behavior.


    Assuming p, n are in [0, INT_BIT)

    unsigned mask = (~((~0U) << n)) << p;
    return (x & ~mask) + ((y << p) & mask);
    or
    return x ^ ((x ^ (y << p)) & mask)

    --Jonathan
    Jonathan Lee, May 2, 2010
    #3
  4. Alex Vinokur

    Eric Sosman Guest

    On 5/2/2010 11:39 AM, Ben Bacarisse wrote:
    > [...]
    > ~0 is often signed and negative [...]


    In C, s/often/always/.

    > [...]
    > This goes wrong if there are padding bits, but at least we can check for
    > that (UINT_MAX will be == ~0u if there are none).


    Also if there are twenty. 6.5.3.3p4: "[...] If the promoted
    type is an unsigned type, the expression ~E is equivalent to the
    maximum value representable in that type minus E." As always, the
    settings of any padding bits in the result of ~E (of any arithmetic
    operation) are unspecified.

    --
    Eric Sosman
    lid
    Eric Sosman, May 2, 2010
    #4
  5. Eric Sosman <> writes:

    > On 5/2/2010 11:39 AM, Ben Bacarisse wrote:
    >> [...]
    >> ~0 is often signed and negative [...]

    >
    > In C, s/often/always/.


    I'll take your word for it! I was not sure about ~0 on a 1's complement
    machine that supports negative zero. It's called "negative" but is it
    less than zero for the purposes of a shift operation? I was not sure.

    >> [...]
    >> This goes wrong if there are padding bits, but at least we can check for
    >> that (UINT_MAX will be == ~0u if there are none).

    >
    > Also if there are twenty. 6.5.3.3p4: "[...] If the promoted
    > type is an unsigned type, the expression ~E is equivalent to the
    > maximum value representable in that type minus E." As always, the
    > settings of any padding bits in the result of ~E (of any arithmetic
    > operation) are unspecified.


    Ah, yes, of course. Do you know a good way to determine the width of an
    unsigned type? By "good" I probably mean "other than the obvious"
    iterative one.

    --
    Ben.
    Ben Bacarisse, May 2, 2010
    #5
  6. Alex Vinokur

    Eric Sosman Guest

    On 5/2/2010 2:15 PM, Ben Bacarisse wrote:
    > Eric Sosman<> writes:
    >
    >> On 5/2/2010 11:39 AM, Ben Bacarisse wrote:
    >>> [...]
    >>> ~0 is often signed and negative [...]

    >>
    >> In C, s/often/always/.

    >
    > I'll take your word for it! I was not sure about ~0 on a 1's complement
    > machine that supports negative zero. It's called "negative" but is it
    > less than zero for the purposes of a shift operation? I was not sure.


    Ah! Okay, "negative zero" might not be "negative" (since it
    compares equal to "positive zero"). Point taken.

    >>> [...]
    >>> This goes wrong if there are padding bits, but at least we can check for
    >>> that (UINT_MAX will be == ~0u if there are none).

    >>
    >> Also if there are twenty. 6.5.3.3p4: "[...] If the promoted
    >> type is an unsigned type, the expression ~E is equivalent to the
    >> maximum value representable in that type minus E." As always, the
    >> settings of any padding bits in the result of ~E (of any arithmetic
    >> operation) are unspecified.

    >
    > Ah, yes, of course. Do you know a good way to determine the width of an
    > unsigned type? By "good" I probably mean "other than the obvious"
    > iterative one.


    Hallvard B. Furuseth came up with

    "

    /* Number of bits in inttype_MAX, or in any (1<<k)-1 where
    * 0 <= k < 3.2E+10 */
    #define IMAX_BITS(m) ((m) /((m)%0x3fffffffL+1) /0x3fffffffL %0x3fffffffL
    *30 \
    + (m)%0x3fffffffL /((m)%31+1)/31%31*5 +
    4-12/((m)%31+3))

    Or if you are less paranoid about how large UINTMAX_MAX can get:

    /* Number of bits in inttype_MAX, or in any (1<<k)-1 where 0 <= k < 2040 */
    #define IMAX_BITS(m) ((m)/((m)%255+1) / 255%255*8 + 7-86/((m)%255+12))

    .." (Sorry about the line-wrapping.) Dunno whether you'd deem this
    good, but it's certainly a jaw-dropper.

    --
    Eric Sosman
    lid
    Eric Sosman, May 2, 2010
    #6
  7. Eric Sosman <> writes:

    > On 5/2/2010 2:15 PM, Ben Bacarisse wrote:

    <snip>
    >> Do you know a good way to determine the width of an
    >> unsigned type? By "good" I probably mean "other than the obvious"
    >> iterative one.

    >
    > Hallvard B. Furuseth came up with
    >
    > "
    >
    > /* Number of bits in inttype_MAX, or in any (1<<k)-1 where
    > * 0 <= k < 3.2E+10 */
    > #define IMAX_BITS(m) ((m) /((m)%0x3fffffffL+1) /0x3fffffffL
    > %0x3fffffffL *30 \
    > + (m)%0x3fffffffL /((m)%31+1)/31%31*5 +
    > 4-12/((m)%31+3))
    >
    > Or if you are less paranoid about how large UINTMAX_MAX can get:
    >
    > /* Number of bits in inttype_MAX, or in any (1<<k)-1 where 0 <= k < 2040 */
    > #define IMAX_BITS(m) ((m)/((m)%255+1) / 255%255*8 + 7-86/((m)%255+12))
    >
    > ." (Sorry about the line-wrapping.) Dunno whether you'd deem this
    > good, but it's certainly a jaw-dropper.


    I remember seeing that now... And I was worried about suggesting

    x ^ (x ^ y) & mask
    for
    x & ~mask | y & mask

    !!

    --
    Ben.
    Ben Bacarisse, May 2, 2010
    #7
  8. Ben Bacarisse <> writes:

    > Alex Vinokur <> writes:

    <snip>
    >> setbits(x,p,n,y) returns x with the n bits that begin at position p
    >> set to the rightmost n bits of y, leaving the other bits unchanged.

    ^^^^^^^^^^^^^^^^^^^^^
    I missed this bit of the spec. You need y << p - n + 1 in:

    > unsigned width = CHAR_BIT * sizeof x;
    > unsigned mask = ~0u >> width - n << p - n + 1;
    > return x & ~mask | y & mask;


    return x & ~mask | (y << p - n + 1) & mask;

    but you should also use:

    unsigned mask = ~(~0u << n) << p - n + 1;

    as this does not need the width of the type.

    --
    Ben.
    Ben Bacarisse, May 2, 2010
    #8
  9. Jonathan Lee <> writes:

    > On May 2, 7:46 am, Alex Vinokur <> wrote:
    >> Here is Richard Heathfield's function
    >> ...
    >> setbits(x,p,n,y) returns x with the n bits that begin at position p
    >> set to the rightmost n bits of y, leaving the other bits unchanged.
    >>
    >> It seems that setbits (x, 31, n, y) may produce undefined behavior.

    >
    > Assuming p, n are in [0, INT_BIT)
    >
    > unsigned mask = (~((~0U) << n)) << p;


    This is a better way to make the mask but I think you've altered what p
    means. Alex (based on Richard's code) seems to take p to be the
    position of the most significant bit of those changed.

    It's simpler (and consistent with the problem wording) to interpret p as
    the least significant bit of the changed bits (as you have done) but
    some people might be confused by this change of meaning.

    It easy to switch to the other interpretation of the problem: substitute
    p - n + 1 for p.

    > return (x & ~mask) + ((y << p) & mask);
    > or
    > return x ^ ((x ^ (y << p)) & mask)


    --
    Ben.
    Ben Bacarisse, May 2, 2010
    #9
  10. Alex Vinokur

    Alex Vinokur Guest

    "Alex Vinokur" <> wrote in message
    news:...
    > Hi,
    >
    > Here is Richard Heathfield's function from
    > http://users.powernet.co.uk/eton/kandr2/krx206.html
    >
    > #include <stdio.h>
    > unsigned setbits(unsigned x, int p, int n, unsigned y)
    > {
    > return (x & ((~0 << (p + 1)) | (~(~0 << (p + 1 - n))))) | ((y & ~(~0
    > << n)) << (p + 1 - n));
    > }
    >
    > setbits(x,p,n,y) returns x with the n bits that begin at position p
    > set to the rightmost n bits of y, leaving the other bits unchanged.
    >
    > It seems that setbits (x, 31, n, y) may produce undefined behavior.
    >
    > According to the C++ standard:
    > Behavior of shift operators << and >> is undefined if the right
    > operand is negative, or
    > greater than or equal to the length in bits of the promoted left
    > operand.
    >
    > So, result ((~0 << (p + 1)) may be undefined.


    Replace ((~0 << (p + 1)) with (((~0 << (p)) << 1)

    >
    > Regards,
    >
    > Alex Vinokur
    >
    >
    >
    >
    Alex Vinokur, May 3, 2010
    #10
  11. Alex Vinokur

    Alex Vinokur Guest

    Templated setBits() based on fixed Richard Heathfield's function

    Here is templated setBits() based on fixed Richard Heathfield's
    function in http://users.powernet.co.uk/eton/kandr2/krx206.html

    setbits(x,p,n,y) returns x with the n bits that begin at position p
    set to the rightmost n bits of y, leaving the other bits unchanged.

    template<typename T>
    T setBits(T x, std::size_t p, std::size_t n, T y)
    {
    BOOST_STATIC_ASSERT(std::numeric_limits<T>::is_integer);
    BOOST_STATIC_ASSERT(!std::numeric_limits<T>::is_signed);
    assert(n > 0);
    assert(n < (sizeof(T) * CHAR_BIT));
    assert(p < (sizeof(T) * CHAR_BIT));
    assert(p >= (n - 1));

    const T maxT = ~static_cast<T> (0);
    return (x & ((static_cast<T>(maxT << p) << 1) | ~static_cast<T>(maxT
    << (p + 1 - n))) | static_cast<T>((y & ~static_cast<T>(maxT << n)) <<
    (p + 1 - n)));

    // ----------------------------------
    // const T maxT = ~static_cast<T> (0);
    // const T tmp1 = static_cast<T>(maxT << p);
    // const T tmp2 = tmp1 << 1;
    // const T tmp3 = ~static_cast<T>(maxT << (p + 1 - n));
    // const T tmp4 = ~static_cast<T>(maxT << n);
    // const T tmp5 = static_cast<T>((y & tmp4) << (p + 1 - n));

    // return ( x & (tmp2 | tmp3)) | tmp5;
    // ----------------------------------
    }

    Alex
    Alex Vinokur, May 3, 2010
    #11
  12. Alex Vinokur

    Phil Carmody Guest

    Ben Bacarisse <> writes:
    > Alex Vinokur <> writes:
    >> Here is Richard Heathfield's function from http://users.powernet.co.uk/eton/kandr2/krx206.html
    >>
    >> #include <stdio.h>
    >> unsigned setbits(unsigned x, int p, int n, unsigned y)
    >> {
    >> return (x & ((~0 << (p + 1)) | (~(~0 << (p + 1 - n))))) | ((y & ~(~0
    >> << n)) << (p + 1 - n));
    >> }
    >>
    >> setbits(x,p,n,y) returns x with the n bits that begin at position p
    >> set to the rightmost n bits of y, leaving the other bits unchanged.
    >>
    >> It seems that setbits (x, 31, n, y) may produce undefined behavior.

    >
    > Yes, it does.
    >
    >> According to the C++ standard:

    >
    > K&R is about C so you should probably quote the C standard. The intent
    > is that C and C++ remain synchronised about this sort of thing so won't
    > matter much, but a solution to a K&R exercise must be assumed to be C.
    > As it happens, modern C (C99) has diverged from C++ in the case of left
    > shifting negative numbers but the above is, presumably, not C99.
    >
    >> Behavior of shift operators << and >> is undefined if the right
    >> operand is negative, or
    >> greater than or equal to the length in bits of the promoted left
    >> operand.
    >>
    >> So, result ((~0 << (p + 1)) may be undefined.

    >
    > It can be undefined for other reasons, though none that matter in the
    > context of K&R2. ~0 is often signed and negative in which case it is
    > undefined in C99 but not in C90 or in C++ (at least up to and including
    > 2003). It's safer to shift unsigned types so I'd suggest ~0u.
    >
    > I think it's hard to do this without knowing the width of the type. I'd
    > probably write:


    Modulo caveats, so would I.

    > unsigned width = CHAR_BIT * sizeof x;
    > unsigned mask = ~0u >> width - n << p - n + 1;


    Alas that can't portably inject a zero-length bitstring.

    > return x & ~mask | y & mask;
    >
    > This goes wrong if there are padding bits, but at least we can check for
    > that (UINT_MAX will be == ~0u if there are none).
    >
    > There is a bit-twiddling version that is one operation shorter:
    >
    > return x ^ (x ^ a) & mask;


    One C operation shorter, yes. One instruction deeper (3 rather than 2)
    on architectures sufficiently rich to parallelise the '&'s in the former.

    > but you'd need to justify the "eh?" this might prompt in the reader.


    Anyone who can't read that expression from left to right as
    "change in x the bits that are different between x and a within the mask"
    on the second reading shouldn't be reading code, but should be clicking
    on buttons in some GUI instead. It's a perfectly standard idiom amongst
    those who have used C as a high level assembler in every-tick-counts
    embedded work. Or at least should be.

    Phil
    --
    I find the easiest thing to do is to k/f myself and just troll away
    -- David Melville on r.a.s.f1
    Phil Carmody, May 3, 2010
    #12
  13. Alex Vinokur

    Phil Carmody Guest

    Ben Bacarisse <> writes:
    > context of K&R2. ~0 is often signed and negative in which case it is
    > undefined in C99 but not in C90 or in C++ (at least up to and including
    > 2003). It's safer to shift unsigned types so I'd suggest ~0u.


    ~0u may end up being converted to being signed on sufficiently
    bizarre architectures which permit the usual arithmetic
    conversions (6.3.1.8) to fall through to, and be caught by:

    Otherwise, if the type of the operand with signed
    integer type can represent all of the values of
    the type of the operand with unsigned integer
    type, then the operand with unsigned integer type
    is converted to the type of the operand with
    signed integer type.

    But that probably only affects people in 33-bit la-la-land.

    Phil
    --
    I find the easiest thing to do is to k/f myself and just troll away
    -- David Melville on r.a.s.f1
    Phil Carmody, May 3, 2010
    #13
  14. Phil Carmody <> writes:

    > Ben Bacarisse <> writes:
    >> Alex Vinokur <> writes:

    <snip>
    >>> setbits(x,p,n,y) returns x with the n bits that begin at position p
    >>> set to the rightmost n bits of y, leaving the other bits unchanged.

    <snip>
    >> unsigned width = CHAR_BIT * sizeof x;
    >> unsigned mask = ~0u >> width - n << p - n + 1;

    >
    > Alas that can't portably inject a zero-length bitstring.


    True. I don't know if that matters or not (the specification is a
    little loose) but I have already noted that JL's construction of the
    mask is superior and, since it handles 0 naturally, it wins all round:

    unsigned width = ~(~0u << n) << p - n + 1;
    return x & ~mask | (y << p - n + 1) & mask;

    If one goes on to interpret p as the least significant bit of those
    injected then there is no need for p - n + 1; and the solution is
    simpler still:

    unsigned mask = ~(~0u << n) << p;
    return x & ~mask | (y << p) & mask;

    which is what he posted (modulo parentheses).

    [I'm posting just to summarise what I think is the neatest solution.]

    <snip>
    --
    Ben.
    Ben Bacarisse, May 3, 2010
    #14
    1. Advertising

Want to reply to this thread or ask your own question?

It takes just 2 minutes to sign up (and it's free!). Just click the sign up button to choose a username and then you can ask your own questions on the forum.
Similar Threads
  1. Replies:
    12
    Views:
    632
  2. Malcolm McLean

    Richard Heathfield's personal threads.

    Malcolm McLean, Nov 9, 2007, in forum: C Programming
    Replies:
    52
    Views:
    1,174
    Christopher Benson-Manica
    Nov 21, 2007
  3. Tomás Ó hÉilidhe

    [OFF-TOPIC] Animosity on the part of Mr Richard Heathfield

    Tomás Ó hÉilidhe, Dec 13, 2008, in forum: C Programming
    Replies:
    113
    Views:
    1,970
    Richard Bos
    Dec 22, 2008
  4. Anonymous

    Animosity on the part of Mr Richard Heathfield

    Anonymous, Dec 14, 2008, in forum: C Programming
    Replies:
    6
    Views:
    386
    CBFalconer
    Dec 21, 2008
  5. Alex Vinokur

    Richard Heathfield's setbits()

    Alex Vinokur, May 2, 2010, in forum: C Programming
    Replies:
    25
    Views:
    734
    Tim Rentsch
    May 6, 2010
Loading...

Share This Page